Question

A diving bell in the shape of a cylinder with a height of 2.4m closed at the upper end and open at the lower end. The bell lowered from air into the sea water of density=1,020 kg/m3. The air in the bell is initially at 20 Celsius. The bell lowered to a depth (measured to the bottom of the bell) of 70m. At this depth the water temperature is 5 Celsius, and the bell is in thermal equilibrium with the water. How high does sea water rise in the bell? Give your answer in m to the nearest 0.01m.

take atmospheric pressure to be P0= 101325 Pa, the gravitational constant to be g=9.8 m/s2, and absolute zero to be at -273 Celsius.

Answer 1

According to combined gas law:

$$\frac{P_0V_0}{T_0}=\frac{P_1V_1}{T_1},\\\space\\ \frac{P_0\pi r^2h}{T_0}=\frac{(P_0+\rho gD)V_1}{T_1}.\\\space\\ V_1=\frac{T_1P_0\pi r^2h}{T_0(P_0+\rho gD)}.$$

The height of the air column in the bell is

$$y=\frac{V_1}{\pi r^2}=\frac{T_1P_0h}{T_0(P_0+\rho gD)}=0.29\text{ m}.$$

The height of water is

$$w=h-y=2.11\text{ m}.$$